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2 years ago

Let P(x) denote the statement "2x+5 > 10." Which of the following is true?

Mathematics

1 answer:

ruslelena [56]2 years ago

4 0

Answer: P(3) is True

Step-by-step explanation:

The given statement is an inequality denoted as P(x). To find out which of the options is true you have to evaluate each given value of X in the inequality and perform the arithmetic operations, then you have to see if the expression makes sense.

For P(0): Replace X=0 in 2x+5>10

2(0)+5>10

0+5>10

5>10 is false because 5 is not greater than 10

For P(3): Replace X=3 in 2x+5>10

2(3)+5>10

6+5>10

11>10 is true because 11 is greater than 10

For P(2): Replace X=2 in 2x+5>10

2(2)+5>10

4+5>10

9>10 is false

For P(1): Replace X=1 in 2x+5>10

2(1)+5>10

2+5>10

7>10 is false

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The length of a rectangle is 4 less than three times it’s width. The perimeter is 72 feet. Find the dimensions of a rectangle

ratelena [41]

Answer:

Step-by-step explanation:

so let’s say x is the length. Then the width will be 3x-4. You add them together and multiply by two. The perimeter is 72 so the equation equala 72.

Equation:

(x+3x-4)*2=72

(4x-4)*2=72

8x-8=72

8x=80

x=10

length:10

width:26

6 0

3 years ago

Premier Bank is planning to have a new building constructed. They would like the length of the building to be 40 feet longer tha

MaRussiya [10]

Let L, W, F, H, P represent, respectively, the length, width, number of floors, height, and perimeter of the building. Here are some equations that model the given information.

L = W + 40
H = 12×F
foundation cost = 10×L×W
cost of floors = 35000×F
total cost = foundation cost + cost of floors
total cost = 1,180,000
P = 2×(L+W)
P = 3×H

We can rewrite these into a system of equations involving L, W, F.

Let's look first at the relation between the perimeter and height. The two versions of perimeter are equal to each other, so we have (after substituting for H) ...

... 3H = 2(L+W)

... 3(12F) = 2((W +40) +W) . . . . substitute for H and L

... 36F = 4W +80 . . . . . . . . . . . simplify

... 9F = W +20 . . . . . . . . . . . . . divide by 4

Now, we can look at the cost equations.

... 1180000 = 35000F + 10W(W+40) . . . . . substitute for the various cost terms and for L

... 118000 = 3500F +W² +40W . . . . . . . . . divide by 10

The two simplified equations in W and F are ...

9F = W +20
3500F = 118000 -40W -W²

You can solve the first for F and substitute for F in the second. Then you have a quadratic equation in W that can be solved by the usual methods. (Likely, the quadratic formula would be the best choice.) Here, we have elected to let a graphing calculator show the two solutions. One solution has negative dimensions, so is clearly infeasible.

There is one (nearly feasible) solution:

... W ≈ 180.788

... F ≈ 22.31

There is no combination of floor dimensions and integral numbers of floors that will match all of the problem requirements.

_____

A solution that comes about as close as possible is ...